One second after the projection, a stone moves at an angle of ` 45^@` with the horizontal. Two seconds from the start,
it is travelling horizontally. Find the angle of projection with the horizontal. `(g=10 ms^(-2))`.

Options

A. `60^(@)`
B. `tan^(-1)(4)`
C. `tan^(-1)(3)`
D. `tan^(-1)(2)`

Answer

To solve the problem, we need to analyze the motion of the stone in two parts: when it is at an angle of 45 degrees after one second and when it is moving horizontally after two seconds.

### Step 1: Understand the motion of the stone
The stone is projected at an angle θ with the horizontal. After 1 second, it makes an angle of 45 degrees with the horizontal, and after 2 seconds, it is moving horizontally.

### Step 2: Determine the time of flight
Since the stone is moving horizontally at 2 seconds, this indicates that it has reached its maximum height at that point. The time to reach maximum height is half of the total time of flight. Therefore, the total time of flight (T) is 2 seconds, and the time to reach the maximum height is:
\[ t = \frac{T}{2} = 1 \text{ second} \]

### Step 3: Analyze the vertical motion
At the maximum height, the vertical component of the velocity (Vy) becomes zero. We can use the equation of motion:
\[ V_y = u_y - g \cdot t \]
Where:
- \( V_y = 0 \) (at maximum height)
- \( u_y \) = initial vertical component of velocity
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( t = 2 \, \text{s} \) (time to reach the maximum height)

Substituting the values:
\[ 0 = u_y - 10 \cdot 2 \]
\[ u_y = 20 \, \text{m/s} \]

### Step 4: Analyze the horizontal motion
At 2 seconds, the stone is moving horizontally, which means the horizontal component of the velocity (Vx) remains constant throughout the motion. The horizontal component of the initial velocity is:
\[ u_x = V_x \]

### Step 5: Relate the components using the angle
At 1 second, the stone is at a 45-degree angle. Therefore, the vertical and horizontal components of the velocity are equal:
\[ V_y = V_x \]
Since we found \( V_y \) at 1 second:
\[ V_y = 20 \, \text{m/s} \]
Thus,
\[ V_x = 20 \, \text{m/s} \]

### Step 6: Calculate the angle of projection
The angle of projection (θ) can be found using the relationship between the vertical and horizontal components:
\[ \tan(\theta) = \frac{u_y}{u_x} \]
Substituting the values we found:
\[ \tan(\theta) = \frac{20}{20} = 1 \]
Thus,
\[ \theta = \tan^{-1}(1) = 45^\circ \]

### Conclusion
The angle of projection with the horizontal is:
\[ \theta = 45^\circ \]